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Give step-by-step solution with explanation and final answer: add explanations to every stepQ (3). A double window called thermopane is one in which two layers of glass are used | separated by a layer of dry stagnant air. In a given window, each of the glass layers is 6.5 mm thick separated by a 6 mm space of stagnant air. The thermal conductivity of the glass is 0.869 W/m ‘K and that of air is 0.026 over the temperature range used. For a temperature drop of 31 K over the system, calculate the heat loss for a window 1.0 m x 2.0 m.

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Give step-by-step solution with explanation and final answer: add explanations to every stepUploaded ImageQ (3). A double window called thermopane is one in which two layers of glass are used | separated by a layer of dry stagnant air. In a given window, each of the glass layers is 6.5 mm thick separated by a 6 mm space of stagnant air. The thermal conductivity of the glass is 0.869 W/m ‘K and that of air is 0.026 over the temperature range used. For a temperature drop of 31 K over the system, calculate the heat loss for a window 1.0 m x 2.0 m.

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Asked by: Prahlad Kumar
Created at: 2025-09-23 18:14:41

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Let's solve the problem step by step, with explanations for each step. --- ## **Step 1: Understanding the Problem** We have: - Two glass layers, each **6.5 mm** thick - Air gap between glass layers: **6 mm** - **Thermal conductivity (k)**: - Glass: **0.869 W/m·K** - Air: **0.026 W/m·K** - **Temperature difference (\(\Delta T\))**: **31 K** - **Window dimensions**: **1.0 m × 2.0 m** We are to **calculate the total heat loss (Q) through the window**. --- ## **Step 2: Calculate Thermal Resistances** For heat transfer through a flat wall (or window), **thermal resistance \( R \) is:** \[ R = \frac{L}{kA} \] Where: - \( L \) = thickness (m) - \( k \) = thermal conductivity (W/m·K) - \( A \) = area (m²) **Glass Layer:** - Thickness = 6.5 mm = 0.0065 m - \( k_{glass} = 0.869 \) W/m·K **Air Layer:** - Thickness = 6 mm = 0.006 m - \( k_{air} = 0.026 \) W/m·K **Window Area:** \[ A = 1.0 \times 2.0 = 2.0 \text{ m}^2 \] Now, calculate the resistance for each layer: ### **a) Resistance of One Glass Layer** \[ R_{glass} = \frac{L_{glass}}{k_{glass} \cdot A} = \frac{0.0065}{0.869 \times 2} \] \[ R_{glass} = \frac{0.0065}{1.738} \approx 0.00374 \text{ K/W} \] ### **b) Resistance of Air Layer** \[ R_{air} = \frac{L_{air}}{k_{air} \cdot A} = \frac{0.006}{0.026 \times 2} \] \[ R_{air} = \frac{0.006}{0.052} \approx 0.1154 \text{ K/W} \] --- ## **Step 3: Total Thermal Resistance** Since the layers are **in series**, total resistance is the sum: \[ R_{total} = R_{glass1} + R_{air} + R_{glass2} \] \[ R_{total} = 0.00374 + 0.1154 + 0.00374 = 0.12288 \text{ K/W} \] --- ## **Step 4: Calculate Heat Loss** The heat loss rate \( Q \) is given by: \[ Q = \frac{\Delta T}{R_{total}} \] \[ Q = \frac{31}{0.12288} \approx 252.25 \text{ W} \] --- ## **Step 5: Final Answer** ### **The heat loss through the window is:** \[ \boxed{252 \text{ W}} \] --- ### **Summary Table of Resistances** | Layer | Resistance (K/W) | |---------------|------------------| | Glass (1) | 0.00374 | | Air | 0.1154 | | Glass (2) | 0.00374 | | **Total** | **0.12288** | --- **Explanation:** The heat transfer is limited mainly by the air gap, which has much higher resistance than the glass. The total resistance allows you to use the temperature difference to find the heat loss using Ohm's Law analogy for heat transfer.

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